Question

find dy/dx using logarithmic differentiation: y=2x^ln2x Click to see my work so far...

Answer 1

lny=ln2x^ln2x --> lny=(ln2x)(ln2x) --> (1/y)(dy/dx)=(ln2x)(1/2x) + (ln2x)(1/2x) --> dy/dx=y(ln2x/2x)+(ln2x/2x) --> ?

Answer 2

hmmm ca you make the (ln2x)(ln2x) = (ln2x)^2 and then take the derivative of that?

Answer 3

you're missing the last bit of the chain rules out here.

to make it clearer, \((\ln 2 x)' = {d \ln (u)\over du} { du\over dx}\), where \(u = 2 x\)

that \(\frac{d(2x)}{dx}\) bit is missing

personally i'd do the RHS as all chain rule: \((\ln ^2 2x)' = 2 \ln 2x . \frac{1}{2x}.2\). looks really simple that way , yes?

:p

Answer 4

\[y=2x^{\ln2x}~\\~lny=\ln(2x) \cdot~\ln(2x)=(\ln(2x))^2\]

Answer 5

\[\frac{1}{y}\frac{dy}{dx}=\frac{2\ln(2x)}{2x}~\cdot~2\]
correct?

Answer 6

@Amenah8 u der?
u can also use product rule
i just simplified the work

Answer 7

@Amenah8 m i clear to u?

Answer 8

sorry! i was in class. yes, that really helped thank you so much